A) \[\frac{-10}{7}\]
B) \[\frac{10}{7}\]
C) \[\frac{-10}{11}\]
D) \[\frac{10}{11}\]
E) \[\frac{10}{9}\]
Correct Answer: A
Solution :
Given lines can be rewritten as \[\frac{x-1}{-3}=\frac{y-2}{2\alpha }=\frac{z-3}{2}\] and \[\frac{x-1}{3\alpha }=\frac{y-1}{1}=\frac{z-6}{-5}\] Since, lines are perpendicular. \[\therefore \] \[{{a}_{1}}{{a}_{2}}+b{{ & }_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0\] \[\Rightarrow \] \[(-3)(3\alpha )+2\alpha (1)+(2)-5=0\] \[\Rightarrow \] \[-9\alpha +2\alpha -10=0\] \[\Rightarrow \] \[\alpha =-\frac{10}{7}\]
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